Question

我将尽可能清楚地了解这个问题，但如果您需要更多信息，请询问。我有一个tableviewcontroller，其中包含登录用户与该应用程序的其他用户共享的所有消息的列表。当点击登录用户点击一个单元格时，我希望用户能够转到一个视图控制器，允许他们与他们喜欢的任何用户聊天。这个聊天是使用JSQMessageController获得的。但是，当我在tableviewcontroller中设置segue时，显示如下：

 override func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {

             let message = messages[indexPath.row]

            if message.ReceiverId != self.loggedInUserUid {
                var newVariable = message.ReceiverId

                if self.userpicuid ==  newVariable {

                    let ref = FIRDatabase.database().reference().child("users").child(userpicuid!)

                    ref.observeSingleEvent(of: .value, with: { (snapshot)
                        in

                        if let dictionary = snapshot.value as? [String: AnyObject]{
                            for post in dictionary {
                                let messages = post.value as! [String: AnyObject]
                                for (id, value) in messages {

                                    self.username = messages["username"] as? String


                                }}}})}} else if message.senderId != self.loggedInUserUid {
                let newVariable = message.senderId

                if self.userpicuid ==  newVariable {

                    let ref = FIRDatabase.database().reference().child("users").child(userpicuid!)

                    ref.observeSingleEvent(of: .value, with: { (snapshot)
                        in

                        if let dictionary = snapshot.value as? [String: AnyObject]{
                            for post in dictionary {
                                let messages = post.value as! [String: AnyObject]
                                for (id, value) in messages {

                                    self.username = messages["username"] as? String
                                }}}})}

            }

        performSegue(withIdentifier: "MessageNow", sender: self.userpicuid)

    }


    override public func prepare(for segue: UIStoryboardSegue, sender: Any?) {
        guard segue.identifier == "MessageNow", let chatVc = segue.destination as? SendMessageViewController else {
            return
        }

        chatVc.senderId = self.loggedInUser?.uid
        chatVc.receiverData = sender as AnyObject
        chatVc.senderDisplayName = self.userpicuid
        chatVc.username = self.username

    }

我在MessageViewController中收到错误：

 var receiverData: AnyObject?

   override func viewDidLoad() {
        super.viewDidLoad()

let receiverId = receiverData as! String
        let receiverIdFive = String(receiverId.characters.prefix(5))
        let senderIdFive = String(senderId.characters.prefix(5))
        if (senderIdFive > receiverIdFive)
        {
            self.convoId = senderIdFive + receiverIdFive
        }
        else
        {
            self.convoId = receiverIdFive + senderIdFive
        }}

我在let receiverId = receiverData上得到了错误！字符串： 无法转换类型＆＃39; Chat_App.MessageTableViewCell＆＃39; （0x10eb2ef10）到＆＃39; NSString＆＃39; （0x110ab1c60）。

在另一个视图控制器中，我有：

 @IBAction func sendMessage(_ sender: Any) {

    performSegue(withIdentifier: "sendMessageToUser", sender: self.userpicuid)
}

override public func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    guard segue.identifier == "sendMessageToUser", let chatVc = segue.destination as? SendMessageViewController else {
        return
    }

    chatVc.senderId = self.loggedInUser?.uid
    chatVc.receiverData = sender as! String!
    chatVc.senderDisplayName = self.userpicuid
    chatVc.username = self.username

}

它非常完美。

Answer 1

您的sender是发起segue的UITableViewCell。将其强制转换为NSString时崩溃。删除此行

let receiverId = receiverData as! String

在prepareForSegue执行此操作

chatVc.receiverData = self.userpicuid

Answer 2

发件人是谁？并且你将它转换为AnyObject。 AnyObject引用类类型，并在您尝试将其强制转换为Swift值类型（String）时进行断言。

请改为尝试：

chatVc.receiverData = NSString(string: sender as! String)

使用tableViewCells

2 个答案: